Question 5In a certain school district, it was observed that 30% of the students in the element schools were classified asonly children (no siblings). However, in the special program for talented and gifted children, 81 out of 209students are only children. The school district administrators want to know if the proportion of only childrenin the special program is different from the proportion for the school district. Test at the a =significance.0.02 level ofWhat is the hypothesized population proportion for this test?p =(Report answer as a decimal accurate to 2 decimal places. Do not report using the percent symbol.)Based on the statement of this problem, how many tails would this hypothesis test have?one-tailed testtwo-tailed testChoose the correct pair of hypotheses for this situation:(A)(B)(C)Ho:p = 0.3Ha:p 0.30.3(D)(E)(F)Но: р — 0.388 | Но:р — 0.388 | Но : р — 0.388Ha:p 0.388(A) (B) (C) (D) (E) (F)Using the normal approximation for the binomial distribution, what is the test statistic for this sample basedon the sample proportion?= Z(Report answer as a decimal accurate to 3 decimal places.)You are now ready to calculate the p-value for this sample.p-value :%3D(Report answer as a decimal accurate to 4 decimal places.)This p-value (and test statistic) leads to a decision to (А)(B)(C)Ho:p = 0.3Ha:p 0.3(D)(E)(F)Но:р — 0.388| Но:рHa:p 0.3880.388 Ho:p = 0.388(A) (B) (C) (D) (E) (F)Using the normal approximation for the binomial distribution, what is the test statistic for this sample basedon the sample proportion?z =(Report answer as a decimal accurate to 3 decimal places.)You are now ready to calculate the p-value for this sample.p-value(Report answer as a decimal accurate to 4 decimal places.)This p-value (and test statistic) leads to a decision to...O reject the null hypothesisO accept the null hypothesisfail to reject the null hypothesisreject the alternative hypothesisAs such, the final conclusion is that...There is enough evidence to show that there is not a different proportion of only children in the G&Tprogram.There is not enough evidence to show that there is not a different proportion of only children in theG&T program.There is enough evidence to show that there is a different proportion of only children in the G&Tprogram.There is not enough evidence to show that there is a different proportion of only children in the G&Tprogram.

Name: Question 5>In a certain school district, it was observed that 30% of
Uploaded: 2024-03-20T07:07:00.000Z
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Description: Question 5>In a certain school district, it was observed that 30% of the students in the element schools were classified asonly children (no siblings). However, in the special program for talented and gifted children, 81 out of 209students are only

Question

Question 5>In a certain school district, it was observed that 30% of the students in the element schools were classified asonly children (no siblings). However, in the special program for talented and gifted children, 81 out of 209students are only children. The school district administrators want to know if the proportion of only childrenin the special program is different from the proportion for the school district. Test at the a =significance.0.02 level ofWhat is the hypothesized population proportion for this test?p =(Report answer as a decimal accurate to 2 decimal places. Do not report using the percent symbol.)Based on the statement of this problem, how many tails would this hypothesis test have?one-tailed testtwo-tailed testChoose the correct pair of hypotheses for this situation:(A)(B)(C)Ho:p = 0.3Ha:p < 0.3Ho :pHa:p # 0.3Но: р — 0.3Ha:p > 0.30.3(D)(E)(F)Но: р — 0.388 | Но:р — 0.388 | Но : р — 0.388Ha:p < 0.388%3DHa:p + 0.388 Ha:p > 0.388(A) (B) (C) (D) (E) (F)Using the normal approximation for the binomial distribution, what is the test statistic for this sample basedon the sample proportion?= Z(Report answer as a decimal accurate to 3 decimal places.)You are now ready to calculate the p-value for this sample.p-value :%3D(Report answer as a decimal accurate to 4 decimal places.)This p-value (and test statistic) leads to a decision to (А)(B)(C)Ho:p = 0.3Ha:p < 0.3Но: р — 0.3Ha:p + 0.3Ho :p = 0.3На:р> 0.3(D)(E)(F)Но:р — 0.388| Но:рHa:p < 0.388 H:p + 0.388 H:p > 0.3880.388 Ho:p = 0.388(A) (B) (C) (D) (E) (F)Using the normal approximation for the binomial distribution, what is the test statistic for this sample basedon the sample proportion?z =(Report answer as a decimal accurate to 3 decimal places.)You are now ready to calculate the p-value for this sample.p-value(Report answer as a decimal accurate to 4 decimal places.)This p-value (and test statistic) leads to a decision to...O reject the null hypothesisO accept the null hypothesisfail to reject the null hypothesisreject the alternative hypothesisAs such, the final conclusion is that...There is enough evidence to show that there is not a different proportion of only children in the G&Tprogram.There is not enough evidence to show that there is not a different proportion of only children in theG&T program.There is enough evidence to show that there is a different proportion of only children in the G&Tprogram.There is not enough evidence to show that there is a different proportion of only children in the G&Tprogram.

Accepted Answer

Let p be the population proportion of students in the element schools were classified as only…